1. Understand the motion

A disk $B$ of radius $R$ rolls बिना slipping on the fixed disk $A$ of the same radius $R$.

Since both radii are $R$, the center of disk $B$ moves in a horizontal circle of radius $2R$ about the center of disk $A$.

Its center has angular speed $\omega$, so the speed of the center of mass of $B$ is $v_{CM}=2R\omega.$

2. Use rolling without slipping to find spin angular speed of disk $B$

At the contact point, the instantaneous velocity must be zero.

So, for disk $B$, $v_{CM}=R\,\Omega,$ where $\Omega$ is the angular speed of spinning of disk $B$ about its own central axis (which is perpendicular to its plane, i.e. vertical).

Thus, $\Omega=\frac{v_{CM}}{R}=\frac{2R\omega}{R}=2\omega.$

3. Angular momentum of disk $B$ about center of $A$

Angular momentum about the center of $A$ has two parts:

$\vec L=\vec L_{orb}+\vec L_{spin}.$

(a) Orbital angular momentum

Treat the center of mass as a particle of mass $M$ moving in a circle of radius $2R$ with angular speed $\omega$.

So, $L_{orb}=M(2R)^2\omega=4M\omega R^2.$

(b) Spin angular momentum

For a thin uniform disk about its own symmetry axis, $I_{CM}=\frac12 MR^2.$

Hence, $L_{spin}=I_{CM}\Omega=\frac12 MR^2(2\omega)=M\omega R^2.$

4. Total angular momentum

Therefore, $L=4M\omega R^2+M\omega R^2=5M\omega R^2.$

Comparing with $L=nM\omega R^2,$ we get $n=5.$

5. Option check

• A: $2$ ❌
• B: $5$ ✅
• C: $\frac72$ ❌
• D: $\frac92$ ❌

So the correct option is B.